Distance between two points

Before we discuss the procedure of finding the distance between points, we shall first throw some light on the position of two points with respect to a given line. These concepts lay the foundation for various other concepts and hence are extremely useful for attempting numerical.

Consider a point (x₁, y₁) and a line ax + by + c = 0. Then, if ax₁ + by₁+ c is of the same sign as ‘c’, then the point (x₁, y₁) lie on the same origin side of ax + by + c = 0 . Conversely, if the sign of c and ax₁ + by₁+ c is opposite, then the point (x₁, y₁) will lie on the non-origin side of ax + by + c = 0.

Let the line be ax + by + c = 0 and P(x₁, y₁), Q(x₂, y₂) be two points. Then, we have two cases:

Case 1:

If P(x₁, y₁) and Q(x₂, y₂) are on the opposite sides of the line ax + by + c = 0, then the point R on the line ax + by + c = 0 divides the line PQ internally in the ratio m₁ : m₂, where m₁/m₂ must be positive.

Co-ordinates of R are {(m₁x₂ + m₂x₁)/(m₁ + m₂₎, (m₁y₂ + m₂y₁)/(m₁ + m₂)}.

Point R lies on the line ax + by + c = 0.

⇒ m₁/m₂ = (ax₁ + by₁ + c)/(ax₂ + by₂ + c) > 0

So that ax₁ + by₁ + c and ax₂ + by₂ + c should have opposite signs.

Case 2:

If ax₁ + by₁ + c and ax₂ + by₂ + c have the same signs then m₁/m₂ = –ve, so that the point R on the line ax + by + c = 0 will divide the line PQ externally in the ratio m₁ : m₂ and the points P(x₁, y₁) and Q(x₂, y₂) are on the same side of the line ax + by + c = 0.

Illustration:

Find the range of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0.

Solution:

As discussed above, we know that ax₁ + by₁ + c and ax₂ + by₂ + c both must be of the same signs for the points to be on the same side of the line.

Now, 3 + 5 – 1 = 7 > 0 ⇒ sin θ + cos θ – 1 > 0

⇒ sin(π/4 + θ) > 1/√2

⇒ π/4 < π/4 + θ < 3π/4

⇒ 0 < θ < π/2.

Hence, this is the required range of θ in the interval (0, π) such that the points (3, 5) and (sin θ, cos θ) lie on the same side of the line x + y – 1 = 0.

Illustration:

Find α, if (α, α²) lies inside the triangle having sides along the lines 2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.

Solution:

Let A, B, C be vertices of the triangle.

A ≡ (–7, 5), B ≡ (5/4, 7/8) and C ≡ (1/3, 1/9).

Sign of A w.r.t. BC is –ve.

If P lies inside the ?ABC, then sign of P will be the same as sign of A w.r.t. the line BC

⇒ 5α – 6α² – 1 < 0.                     …… (1)

Similarly 2α + 3α² – 1 > 0.          …… (2)

And, α + 2α² – 3 < 0.                  …… (3)

Solving, (1), (2) and (3) for α and then taking intersection,

We get α ∈ (1/2, 1) ∪ (–3/2, –1).

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Illustration:

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, –2), find the equation of BC.

Solution:

From the figure, since OE and OF are the perpendicular bisectors of the sides AB and AC, so we have the coordinates of points E and F as

E ≡ (x₁ + 1/2, y₁ - 2/2),

F ≡ (x₂ + 1/2, y₂ - 2/2).

Since E and F lie on OE and OF respectively,

x₁ – y₁ + 13 = 0                                  … (1)

and x₂ + 2y₂ – 3 = 0                           … (2)

Also, slope of AB = –1 and slope of AC is 2, so that

x₁ + y₁ + 1 = 0.                                   … (3)

And 2x₂ – y₂ – 4 = 0                           … (4)

Solving these equations, we get the co-ordinates of B and C as

B ≡ (–7, 6) and C ≡ (11/5, 2/5)

⇒ Equation of BC is 14x + 23y – 40 = 0.

Illustration:

Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Two variable points A’ and B’ are taken on the same axes such that OA’ + OB’ = OA + OB. Find the locus of the point of intersection of AB’ and A’B.

Solution:

Let A ≡ (a, 0), B (0, b), A’ ≡ (a’, 0), B’ ≡ (0, b’).

Equation of A’B is x/a' + y/b' = 1.                                          …. (1)

and equation of AB’ is x/a + y/b'  = 1.                                   …. (2)

Subtracting (1) from (2), we get, x (1/a – 1/a') + y(1/b' – 1/b) = 0.

⇒ x(a'–a)/aa' + y(b–b')/bb' = 0.                   [Using a’ – a = b – b’]

⇒ x/a(b–b'+a) + y/bb' = 0 

⇒ b’ = a(a+b)y/ay–bx.                   ….. (3)

From (2), b’x + ay = ab’   ….. (4) we get x + y = a + b which is the required locus.

The distance between two points P(x₁, y₁) and Q(x₂, y₂) is (see the figure given below).

Length PQ = √(x₂ – x₁)² + (y₂ – y₁)²

Proof:

Let P(x₁, y₁) and Q(x₂, y₂) be the two points and let the distance between them be ‘d’. Draw PA, QR parallel to y-axis and PR parallel to x-axis. 

Angle QRP = 90^o  

⇒ d² = PR² + RQ² 

⇒ d² = (x₂ – x₁)² + (y₂ – y₁)²

⇒ d = √(x₂ – x₁)² + (y₂ – y₁)². 

You may also view the video on distance formula

Similarly, in three dimensions, we shall have a third coordinate and the formula in that case changes a bit:

Distance between P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by the formula d = √(x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)².

Let us say we want to know the co-ordinates of point which divides a line segment between two points A(x₁, y₁) and B(x₂, y₂) in the ratio m : n. Then two types of division are possible:

The coordinates of such a point are given by:

Internal Division: {(nx₁ + mx₂)/(m + n), (ny₁ + my₂)/(m + n)}

External Division:  {(mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n)}

Note: 

This is called section formula. 

Let P divide the line segment AB in the ratio m : n. If P is inside AB then it is called internal division; if it is outside AB then it is called external division. 

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n.  

Proof: 

Consider ? ABB’  

Since BB’||PQ and AP:PB = m:n (see figure given below)

AQ/AB' = PQ/BB' = m/(m+n) (= AP/AB) 

⇒ x – x₁/x₂ – x₁ = m/(m+n) 

⇒ x = (nx₁ + mx₂)/(m + n) and y = (ny₁ + my₂)/(m + n) 

If P is outside AB (less assume it is at P’) 

We have (x – x₁)/(x₂ – x₁) = m/(m + n) 

⇒ x = (nx₁ + mx₂)/(m + n) and y = (ny₁ + my₂)/(m + n) 

Similarly if P is at P” then 

x = (–mx₂ + nx₁)/(n-m), y = (– my₂ + ny₁)/(n-m)

Note:  

m:n can be written as m/n or λ:1. So any point on line joining A and B will be P{(λx² + x₁)/(λ + 1), (λy² + y₁)/(λ +1)}. It is useful to assume λ:1 because it involves only one variable. 

Illustration: 

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are {(x₁ tan A + x₂ tan B + x₃ tan C)/(tan A + tan B + tan C), (y₁ tan A + y₂ tan B + y₃ tan C)/(tan A + tan B + tan C)}. 

Solution: 

In triangle A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃), draw AD perpendicular to BC. Our effort now should be to find the co-ordinates of the point D.

To do that, we need to find BC/CD. (figure is given above) 

tan B = AD/BD and tan C = AD/CD 

⇒ BD/DC = tan C/tan B 

Now we apply section formulae: 

x_D = (x₂ tan B + x₃ tan C)/(tan B + tan C)                           …… (i)

y_D = (y₂ tan B + y₃ tan C)/(tan B + tan C)                               …… (ii)

We know that orthocenter will lie on AD. We need to find this point and its co-ordinates. 

We should select a point H₁ on AD and take the ratio AH₁/H₁D in such a manner so that x_H1 and y_H1 calculated form (i) should be symmetric in x₁, x₂, x₃, tan A, tan B and tan C.

Let AH₁/H₁D = (tan B + tan C)/tan A 

⇒ xH₁ = (x₁ tan A + x₂ tan B + x₃ tan C)/(tan A + tan B + tan C)

and ⇒ yH₁ = (y₁ tan A + y₂ tan B + y₃ tan C)/(tan A + tan B + tan C) 

Since the result is symmetric, this point H1 will lie on other altitude as well i.e. the altitudes are concurrent

⇒ xH = xH₁ and yH = yH₁.

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Prove analytically that in a right angled triangle the midpoint of the hypotenuse is equidistant from the three angular points. 

Solution: 

While proving a problem analytically take most convenient co-ordinates of known points. 

In the present case triangle is assumed as AOB with coordinates as shown in figure given below, C is midpoint of AB.

So co-ordinates of C will be (a/2, b/2) 

Now AB = √a² + b²

CA = CB = AB/2 (C is mid point of AB) 

= √a² + b² 

And, we know that the distance between two points C and O is given by 

CO = √(a/2 – 0)² + (b/2 – 0)² = √a² + b²/2 

Hence CA = CB = CO 

Coordinates of the point P dividing the join of two points A(x₁, y₁) and B(x₂, y₂) internally in the given ratio λ₁ : λ₂ i.e., AP/BP = λ₁/λ₂ are P {(λ₂x₁ + λ₁x₂)/(λ₂ + λ₁), (λ₂y₁ + λ₁y₂)/(λ₂ + λ₁)}. 

Coordinates of the point P dividing the join of two points A(x₁, y₁) and B(x₂, y₂) externally in the ratio λ₁ : λ₂ i.e., AO/BP = λ₁/λ₂ are P(λ₂x₁+λ₁x₂)/(λ₂–λ₁), (λ₂y₁+λ₁y₂)/(λ₂–λ₁).
 

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